Question

A hydrogen atom, unipositive helium, and dipositive lithium contain a single electron in the same shell. The radius of the shell:
A.Decreases
B.Increases
C.Remains unaffected
D.Cannot be predicted

Answer 1

Hint: For this question we must have the knowledge of the Bohr model for hydrogen and hydrogen-like species. We can calculate the radius of the shell by using the atomic number of their respective species in the formula.

Formula used: \[{{\text{R}}_{\text{n}}} = {{\text{R}}_ \circ } \times \dfrac{{{{\text{n}}^2}}}{{\text{Z}}}\] where \[{{\text{R}}_{\text{n}}}\] is radius of \[{{\text{R}}_{\text{n}}}\] orbital \[{{\text{R}}_{\text{n}}}\] is Bohr radius that is constant, n is number of orbit and s always a natural number and Z is atomic number of element.

Complete step by step solution:
We need to check the variation of radius for the given species. Let us calculate the radius of atom in each case:
The atomic number of hydrogen is 1. Hence \[{\text{Z }} = 1\] and we need to calculate for \[{\text{n }} = 1\]. Putting the values in the formula we will get:
\[{{\text{R}}_{1({\text{H)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{1} = {{\text{R}}_ \circ }\]
The atomic number of Helium is 2. Helium ion is \[{\text{H}}{{\text{e}}^ + }\], but the atomic number remains the same whether it is an ion or atom of the same elements. Hence \[{\text{Z }} = 2\] and we need to calculate for \[{\text{n }} = 1\]. Putting the values in the formula we will get:
\[{{\text{R}}_{1({\text{H}}{{\text{e}}^ + }{\text{)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{2} = \dfrac{{{{\text{R}}_ \circ }}}{2}\]
The atomic number of Lithium is 3. Dipositive Lithium ion is \[{\text{L}}{{\text{i}}^{2 + }}\], but the atomic number remains same whether it is an ion or atom of same elements. Hence \[{\text{Z }} = 3\] and we need to calculate for \[{\text{n }} = 1\]. Putting the values in the formula we will get:
\[{{\text{R}}_{1({\text{L}}{{\text{i}}^{ + 2}}{\text{)}}}} = {{\text{R}}_ \circ } \times \dfrac{{{1^2}}}{3} = \dfrac{{{{\text{R}}_ \circ }}}{3}\]
Hence the radius of \[{\text{H, H}}{{\text{e}}^ + }{\text{and L}}{{\text{i}}^{ + 2}}\] is \[{{\text{R}}_{\text{o}}}{\text{, }}\dfrac{{{{\text{R}}_{\text{o}}}}}{2},{\text{ }}\dfrac{{{{\text{R}}_{\text{o}}}}}{3}\] respectively. Hence, we can clearly see that radius decreases.
The correct option is A.

Note: The given species \[{\text{H, H}}{{\text{e}}^ + }{\text{and L}}{{\text{i}}^{ + 2}}\] are iso-electronic in nature. That is all these species have same number of electron though have different atomic number. Hydrogen has 1 electron and 1 proton. Helium has atomic number 2 and have 2 electrons but in \[{\text{H}}{{\text{e}}^ + }\] has 1 less electron due to positive charge that makes 1 electron in \[{\text{H}}{{\text{e}}^ + }\] and similarly 1 electron is there in \[{\text{L}}{{\text{i}}^{2 + }}\] sue to 2 positive charge.